\newproblem{lay:4_5_31}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.5.31}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Consider finite-dimensionals spaces $V$ and $W$, and a linear transformation $T:V\rightarrow W$. Let $H$ be a nonzero subspace of $V$, and let $T(H)$ be the set
	of images of vectors in $H$. Then $T(H)$ is a subspace of $W$, by Exercise 35 in Section 4.2. Prove that $\dim\{T(H)\}\leq \dim\{H\}$.
}{
  % Solution
	Let $k=\dim\{T(H)\}$. If $k=0$, then $k<\dim\{H\}$. If $k>0$, then $T(H)$ must have a basis formed by $k$ vectors in $T(H)$ of the form $T(\mathbf{v}_1)$, $T(\mathbf{v}_2)$,
	..., $T(\mathbf{v}_k)$. Since the set $\{T(\mathbf{v}_1),T(\mathbf{v}_2),...,T(\mathbf{v}_k)\}$ is linearly independent and $T$ is a linear transformation, then by 
	Exercise 4.3.31 the set $\{\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_k\}$ is also linearly independent in $H$ and consequently $\dim\{H\}\geq k$.
}
\useproblem{lay:4_5_31}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
